A 4.08 g sample of a compound of nitrogen and oxygen contains 3.02 g of oxygen. What is the empirical formula?

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Answer 1 · 2016-07-28T02:03:12

The empirical formula is $N_{4} O$ $"N"_4"O"$ .

An empirical formula is the lowest whole number ratio of elements in a compound. Each ratio will be the subscript of each element.

Determine the mass of each element in grams.

$N :$ $"N":$ $4.08 g - 3.02g = 1.06 g$ $"4.08 g"-"3.02g"="1.06 g"$

$O :$ $"O":$ $3.02 g$ $"3.02 g"$ (given)

Determine moles of each element by dividing the mass of each element by its molar mass (atomic weight on the periodic table in g/mol.

$N :$ $"N":$ $\frac{1.06 g N}{14.007 g/mol} = 0.0756764 mol N$ $"1.06 g N"/"14.007 g/mol"="0.0756764 mol N"$

$O :$ $"O":$ $\frac{3.02 g O}{15.999 g/mol} = 0.188761 mol O$ $"3.02 g O"/"15.999 g/mol"="0.188761 mol O"$

Determine mole ratios by dividing the moles of each element by the least number of moles.

$N :$ $"N":$ $\frac{0.0756764 mol}{0.188761 mol} = 4.01 \approx 4$ $"0.0756764 mol"/"0.188761 mol"="4.01"~~4$

$O :$ $"O":$ $\frac{0.188761 mol}{0.188761} = 1.00 = 1$ $"0.188761 mol"/"0.188761"=1.00=1$

The empirical formula is $N_{4} O$ $"N"_4"O"$ .

The compound with this formula is called nitrosylazide with the following structure.

![https://en.m.wikipedia.org/wiki/Nitrosylazide]( useruploads.socratic.org )

For more examples of this type of empirical formula problem, check out the following website: http://www.chemteam.info/Mole/Emp-formula-given-mass-data.html